Finding the mean from the variance

Question

Is it possible to find the mean from the variance?

Here is the problem:

Supposed that in 3-D $X \sim \mathcal N(\mu, P)$ where $P = 10000 * I_3$. What is the radius of a sphere centered at $\mu$ having a 95% containment probability?

While I understand that the radius of the sphere can be found by: $$R=\sigma*\sqrt{Y} = ||X - \mu||$$ where $Y = (X - \mu)^T P^{-1} (X- \mu)$

Finding that can be done by a nice clean little MATLAB command:

sqrt(chi2inv(0.95,mu))

However, I'm not sure how to obtain $\mu$ in this case. Is it possible to derive $\mu$ from $\sigma$? Or am I overthinking this and this must be solve at a more abstract level? Extra points if there is a nice MATLAB way to go about solving it - especially since I can use the above command to get the radius without hand writing the math.

Thanks for your help in advanced!

Answer 1

You are not even using chi2inv correctly. The second argument is the degrees of freedom of the chi-squared distribution that you want to use, not $\mu$.

You want to find $R$ such that $P(\|X - \mu\|^2 \le R^2) = 0.95$. Since $X \sim N(\mu, 10000 I_3)$ we have $Z \sim N(0, I_3)$ where $Z := (X - \mu) / 100$. Thus we want $R$ such that $P(\|Z\|^2 \le (R / 100)^2) = 0.95$. Since $\|Z\|^2$ follows a chi-squared distribution with $3$ degrees of freedom, you can find the $95\%$ quantile using chi2inv(0.95, 3) and set it equal to $(R/100)^2$ and solve for $R$.